If $D=\{(x,y):x^2+y^2\leq 1\}$. Show there is $p_0\in D$ such that $T(p_0)=(0,0)$

Question

Let $D=\{(x,y):x^2+y^2\leq 1\}$, $T$ transformation of class $C'$ on an open containing $D$, $$T:\left\lbrace \begin{array}{rcl} u &=& f(x,y) \\ v&=&g(x,y) \end{array}\right. $$ whose Jacobian is never $0$ in $D$, and $|T(p)-p|\leq \frac{1}{3}$ for all $p\in D$. Show there is $p_0\in D$ such that $T(p_0)=(0,0)$

Idea: Seems like fixed point theorem can help, for example if I define $H(p)=T(p)-p$, then $H(p)=p$ iff $T(p)=0$, so it's enough to find such point for $H$, by hypothesis $|H(p)|\leq \frac{1}{3}$ . Usually the fixed theorem is stated as: if $f:D\to D$ is continuous then there is a fixed point, or also there is no retraction of class $C'$ from $D$ onto the unit circle (only the boundary of $D$). Here I know that the jacobian of $T$, $J(p)\neq 0$ for all $p\in D$ so there is a neighborhood of any $p\in D$ where $T$ is $1$-to-$1$, i.e., in where $T^{-1}$ exists. But at least for now I can't have a precise idea. Any hints are welcome.

For context, this is a problem #17 section 7.5 from Buck's Advance Calculus. In this section we have the results:

1. For $T:D\subset\mathbb{R}^n \to \mathbb{R}^n$ of class $C'$, $D$ open, with $J(p)\neq 0$ for all $p\in D$ then $T$ is locally $1$-to-$1$ in $D$.
2. If $J(p_0)\neq 0$ only for a specific value, then $T$ is $1$-to-$1$ only in a small neighborhood of $p_0$, where $T^{-1}$ exists.
3. If conditions as in 1.,then $T(D)$ is an open set.

Answer 1

It took me a bit to figure out how to do this without homotopy or winding numbers, which is how it's always proved. This is an interesting exercise; to see how this normally arises, look up the "dog on a leash" or "walking the dog" lemma of complex analysis.

Define the boundary $\partial(A)$ of a set $A \subset \mathbb{R}^n$ to be the intersection $\overline{A} \cap \overline{\mathbb{R}^n \setminus A}$. Thus it's the points which are arbitrarily close to both $A$ and its complement; for example, the unit circle is the boundary of $D$.

Note by definition that if $x \in A$ and $x \notin \partial(A)$, then $x$ is in the interior of $A$ - because if every open neighborhood of $x$ intersected the complement of $A$, then $x$ would be in the boundary. Thus a set is composed of its interior points and its boundary points.

Lemma: If $D \subset \mathbb{R}^n$ and $T$ is $C'$ with non-vanishing Jacobian on (a neighborhood of) $D$ then $T$ doesn't send an interior point of $D$ to the boundary of $T(D)$.

Proof: If $x \in U \subset D$ is a sufficiently small open neighborhood of $x$ in $D$, then $T(U) = V$ is open by the open mapping theorem, and $V = T(U) \subset T(D)$ so that $x$ is an interior point of $T(D)$. QED

Now returning to your problem, if $B$ is the closed ball centered around $0$ with radius $\frac{1}{3}$ then $T(0) \in B$ since $|T(0) - 0| = |T(0)| \leq \frac{1}{3}$. However, $T(\mathbb{S}^1) \cap B = \varnothing$ since $|T(p)-p| \leq \frac{1}{3}$ and thus restricting to $\mathbb{S}^1 = \partial(D)$ gives us points at least $\frac{2}{3}$rds away from $0$. Since $\mathbb{S}^1$ is the boundary of $D$, by the above lemma $T(D)$ has no boundary points in $B$.

Note that $T(D) \cap B$ is non-empty by virtue of containing at least the point $T(0)$. Thus $T(D)$ must contain all of $B$ in order to not have any boundary points there: Otherwise, there would be no point contained in both $\overline{T(D)}$ and $\overline{B \setminus T(D)}$. But then these two sets would be a separation of $B$, which is connected - impossible.

Thus $0 \in B \subset T(D)$, completing the proof.